Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Probabilitas dan Statistika |
| Periode Ujian | : | November 2018 |
| Nomor Soal | : | 12 |
SOAL
Misalkan X merupakan suatu variable acak diskrit dengan fungsi peluang
\(P\left[ {X = x} \right] = \frac{2}{{{3^x}}}\) untuk \(x = 1,2,3,…\)
Berapa peluang bahwa \(X\) adalah genap ?
- \(\frac{1}{4}\)
- \(\frac{2}{7}\)
- \(\frac{1}{3}\)
- \(\frac{2}{3}\)
- \(\frac{3}{4}\)
| Step 1 | Untuk nilai X adalah bilangan genap (2,4,6,..)
\(\Pr [X = x|x\,\,nilai\,genap] = \frac{2}{{{3^2}}} + \frac{2}{{{3^4}}} + \frac{2}{{{3^6}}} + …\)
\(\Pr [X = x|x\,\,nilai\,genap] = 2\left( {\frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ….} \right)\) |
| Step 2 | \(\left( {\frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ….} \right)\, \sim \,\,Deret\,Geometrik\,Tak\,Hingga\,(a = \frac{1}{{{3^2}}},r = \frac{1}{{{3^2}}})\)
\({S_\infty } = \left( {\frac{{\frac{1}{{{3^2}}}}}{{1 – \frac{1}{{{3^2}}}}}} \right)\)
\({S_\infty } = \left( {\frac{{\frac{1}{9}}}{{\frac{8}{9}}}} \right)\)
\({S_\infty } = \frac{1}{8}\) |
| Maka | \(\Pr [X = x|x\,\,nilai\,genap] = 2\left( {\frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} + ….} \right)\)
\(\Pr [X = x|x\,\,nilai\,genap] = 2\left( {\frac{1}{8}} \right)\)
\(\Pr [X = x|x\,\,nilai\,genap] = \frac{1}{4}\) |
| Jawaban | a. \(\frac{1}{4}\) |
